1046. Last Stone Weight

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1046. Last Stone Weight

Description

You are given an array of integers stones where stones[i] is the weight of the i^th stone.

We are playing a game with the stones. On each turn, we choose the heaviest two stones and smash them together. Suppose the heaviest two stones have weights x and y with x <= y. The result of this smash is:

  • If x == y, both stones are destroyed, and
  • If x != y, the stone of weight x is destroyed, and the stone of weight y has new weight y - x.

At the end of the game, there is at most one stone left.

Return the weight of the last remaining stone. If there are no stones left, return 0.

Example 1:

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Input: stones = [2,7,4,1,8,1]
Output: 1
Explanation:
We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of the last stone.

Example 2:

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Input: stones = [1]
Output: 1

Constraints:

  • 1 <= stones.length <= 30
  • 1 <= stones[i] <= 1000

Hints/Notes

  • 2024/12/27
  • priority queue
  • No solution from 0x3F

Solution

Language: C++

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class Solution {
public:
    int lastStoneWeight(vector<int>& stones) {
        priority_queue<int> pq(stones.begin(), stones.end());
        while (pq.size() > 1) {
            int first = pq.top(); pq.pop();
            int two = pq.top(); pq.pop();
            if (first - two > 0) {
                pq.push(first - two);
            }
        }
        return pq.empty() ? 0 : pq.top();
    }
};